To find the points of contact, we’ll use the coordinates derived in the previous lesson: (–a2m/c, a2/c). The required points are (4, -3) and (-4, 3) (Here m = 4/3 and c = ± 25/3). Have a look at the following figure for clarity

Solution Another problem involving direct application of the formula. But in this case, the center of the circle is not at the origin. We’ll use the second equation: \( (y + f) = m(x + g) ± r\sqrt{1+m^2} \)

The slope of the given line is –3/4. So all we have to plug in this value in the slope form of the tangent. We get y = –3x/4 + 3 – 2(–3/4) ± 5\( \sqrt{1+(3/4)^2} \) or 3x + 4y = 43, and 3x + 4y = –7. Therefore, c = 43 or – 7.

That’ll be all for this lesson. Next, I’ll be talking about tangents at a point. See you there!